C2H5OH + 3O2 --> 2CO2 + 3H2O

How many grams of O2 are consumed when 5.86 ml of Ethanol burns completely (the density of C2H5OH is 0.789 g/ml)?

A worked example. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To find out how many grams of O2 are consumed when 5.86 ml of ethanol burns completely, we need to use the given equation and the density of ethanol.

First, we need to find the number of moles of ethanol burned. We can use the formula:

moles = volume (in liters) × density (in g/ml) ÷ molar mass (in g/mol)

The molar mass of C2H5OH (ethanol) is:
C = 12.01 g/mol
H = 1.01 g/mol (there are 6 hydrogens in ethanol)
O = 16.00 g/mol (there is 1 oxygen in ethanol)

Molar mass of ethanol = (2 × C) + (6 × H) + O
= (2 × 12.01) + (6 × 1.01) + 16.00
= 46.07 g/mol

Now, let's calculate the volume of ethanol in liters:
volume (in liters) = volume (in ml) ÷ 1000
= 5.86 ml ÷ 1000
= 0.00586 L

Next, we can calculate the moles of ethanol burned:
moles of ethanol = 0.00586 L × 0.789 g/ml ÷ 46.07 g/mol

Now, to find the moles of O2 consumed, we use the stoichiometric ratio of the balanced equation:
1 mole of ethanol consumes 3 moles of O2

moles of O2 = moles of ethanol × (3 moles of O2 ÷ 1 mole of ethanol)

Finally, we can convert the moles of O2 to grams:
grams of O2 = moles of O2 × molar mass of O2

The molar mass of O2 (oxygen gas) is:
O = 16.00 g/mol (there are 2 oxygens in O2)

Molar mass of O2 = 2 × O
= 2 × 16.00
= 32.00 g/mol

grams of O2 = moles of O2 × 32.00 g/mol

Now, you can plug the calculated values into the formulas and solve for the grams of O2 consumed.